Chapter 1 - Section 1.6 - Equations and Inequalities Involving Absolute Value - 1.6 Assess Your Understanding - Page 133: 35

The solution set is {$-\frac{1}{2},0$}

$|x^2+3x|=|x^2-2x|$ $x^2+3x=x^2-2x$ or $x^2+3x=-1(x^2-2x)$ $x^2+3x-x^2+2x=0$ or $x^2+3x+x^2-2x=0$ $3x+2x=0$ or $2x^2+x=0$ $5x=0$ or $x(2x+1)=0$ $x=0$ or $x=0$ or $x=-\frac{1}{2}$ The solution set is {$-\frac{1}{2},0$}